# Recurrence results for an “on average” measure preserving transformation

I have a finite measure space $$(X, \mathcal{S}, \mu)$$, and a transformation $$f:X\rightarrow X$$ that "preserves measure on average". That is, for $$A \in \mathcal{S}$$ $$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^n \mu(f^k(A)) = \mu(A)$$ I would like to be able to apply some ergodic or recurrence theorem in this situation, but Poincare Recurrence for example requires that the map $$f$$ be measure-preserving, which this "measure-preserving on average" mapping does not satisfy.

I am wondering if anyone can say if there are any results concerning such "measure-preserving on average" maps and any related recurrence or ergodic theorems, or if there is some way I can use say an altered Poincare Recurrence to makes claims about this mapping.

• Your condition is a bit problematic because $f(A)$ may not be a measurable set. It's generally better to write invariance conditions in terms of $f^{-1}$. If a measure satisfies your condition with $f^{-1}$ in place of $f$, it actually is invariant: you can just substitute $f^{-1}(A)$ for $A$ and you see you have the same limit, so your condition implies $\mu(A)=\mu(f^{-1}A)$. – Anthony Quas Dec 7 '20 at 2:36

## 1 Answer

The phrase "a finite measure space $$(X,\mathcal{S},\mu)$$ where $$\mu$$ is the Lebesgue measure" can only reasonably mean that $$\mu$$ is a nonzero multiple of the counting measure, with $$\mathcal{S}$$ being the powerset of $$X$$. At least, we may assume that, if $$A\subsetneq X$$, then $$\mu(A)<\mu(X)$$. Therefore and because the set $$X$$ is finite, $$M:=\max\{\mu(A)\colon A\subsetneq X\}<\mu(X).$$ So, if $$f(X)\ne X$$, then your displayed condition implies $$\mu(X)\le\lim_{n\to\infty}\frac1n\,\sum_{k=1}^n M=M<\mu(X).$$ This contradiction implies that $$f(X)=X$$. Since the set $$X$$ is finite, this means that $$f$$ is a bijection or, in other words, permutation of the set $$X$$.

On the other hand, if every permutation of $$X$$ "preserves a measure $$\mu$$ on average", then it is easy to see that $$\mu$$ must be a multiple of the counting measure.

In particular, it follows that, if $$\mu$$ is a nonzero multiple of the counting measure on $$X$$, then a map $$f\colon X\to X$$ "preserves the measure $$\mu$$ on average" iff $$f$$ is a permutation of the set $$X$$.

• It may be that "measure space with finite measure" is what was intended. Maybe OP can clarify… – Dirk Dec 6 '20 at 19:56
• Yes I meant just measure space of finite measure, I am not sure why I added the Lebesgue measure part. – GEG Dec 6 '20 at 19:57
• I edited the question appropriately, and if this changes much in your answer I would very much appreciate an explanation of what changes. – GEG Dec 6 '20 at 19:59
• @GEG : I think "a finite measure space" can only mean a measure space which is finite, and that is something quite different from "a measure space with a finite measure" or "a finite-measure measure space". Of course, the answer entirely relies on the condition that the space is finite. – Iosif Pinelis Dec 6 '20 at 21:35